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Question
For certain bivariate data the following information is available.
| X | Y | |
| Mean | 13 | 17 |
| S.D. | 3 | 2 |
Correlation coefficient between x and y is 0.6. estimate x when y = 15 and estimate y when x = 10.
Solution
Given, `bar x = 13, bar y = 1, sigma_"X" 3, sigma_"Y" = 2,` r = 0.6
`"b"_"YX" = "r" sigma_"Y"/sigma_"X" = 0.6 xx 2/3 = 0.4`
`"b"_"XY" = "r" sigma_"X"/sigma_"Y" = 0.6 xx 3/2 = 0.9`
The regression equation of X on Y is given by
`("X" - bar x) = "b"_"XY" ("Y" - bar y)`
(X - 13) = 0.9 (Y - 17)
X - 13 = 0.9Y - 15.3
X = 0.9Y - 15.3 + 13
X = - 2.3 + 0.9Y ....(i)
For Y = 15, from equation (i) we get
X = - 2.3 + (0.9)(15) = - 2.3 + 13.5 = 11.2
The regression equation of Y on X is given by
`("Y" - bar y) = "b"_"YX" ("X" - bar x)`
(Y - 17) = 0.4(X - 13)
Y - 17 = 0.4X - 5.2
Y = 0.4X - 5.2 + 17
Y = 11.8 + 0.4X .....(ii)
For X = 10, from equation (ii) we get
Y = 11.8 + 0.4(10) = 11.8 + 4 = 15.8
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| x | y | `x - barx` | `y - bary` | `(x - barx)(y - bary)` | `(x - barx)^2` | `(y - bary)^2` |
| 1 | 5 | – 2 | – 4 | 8 | 4 | 16 |
| 2 | 7 | – 1 | – 2 | `square` | 1 | 4 |
| 3 | 9 | 0 | 0 | 0 | 0 | 0 |
| 4 | 11 | 1 | 2 | 2 | 4 | 4 |
| 5 | 13 | 2 | 4 | 8 | 1 | 16 |
| Total = 15 | Total = 45 | Total = 0 | Total = 0 | Total = `square` | Total = 10 | Total = 40 |
Mean of x = `barx = square`
Mean of y = `bary = square`
bxy = `square/square`
byx = `square/square`
Regression equation of x on y is `(x - barx) = "b"_(xy) (y - bary)`
∴ Regression equation x on y is `square`
Regression equation of y on x is `(y - bary) = "b"_(yx) (x - barx)`
∴ Regression equation of y on x is `square`
